Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $x = \dfrac{r^2 + 2r - 35}{r - 5} \times \dfrac{r - 6}{7r^2 + 49r} $
Solution: First factor the quadratic. $x = \dfrac{(r + 7)(r - 5)}{r - 5} \times \dfrac{r - 6}{7r^2 + 49r} $ Then factor out any other terms. $x = \dfrac{(r + 7)(r - 5)}{r - 5} \times \dfrac{r - 6}{7r(r + 7)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ (r + 7)(r - 5) \times (r - 6) } { (r - 5) \times 7r(r + 7) } $ $x = \dfrac{ (r + 7)(r - 5)(r - 6)}{ 7r(r - 5)(r + 7)} $ Notice that $(r - 5)$ and $(r + 7)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ \cancel{(r + 7)}(r - 5)(r - 6)}{ 7r(r - 5)\cancel{(r + 7)}} $ We are dividing by $r + 7$ , so $r + 7 \neq 0$ Therefore, $r \neq -7$ $x = \dfrac{ \cancel{(r + 7)}\cancel{(r - 5)}(r - 6)}{ 7r\cancel{(r - 5)}\cancel{(r + 7)}} $ We are dividing by $r - 5$ , so $r - 5 \neq 0$ Therefore, $r \neq 5$ $x = \dfrac{r - 6}{7r} ; \space r \neq -7 ; \space r \neq 5 $